# An infinite ODE

Via reddit, a challenge to solve:

$y = y’ + y’’ + y’’’ + \dots$

Once you’ve stopped running away and hiding, I’ll show you the solution they suggested.

The trick is to notice that $y’ = y’’ + y’’’ + y^{(4)} + \dots$, so $y = y’ + (y’)$, which is straightforward to solve:

- $y = 2y’$
- $y = Ae^{x/2}$.

It’s a good idea to check that it works:

- $y’ = \frac{1}{2}y$
- $y’’ = \frac{1}{4}y$
- $\dots$

The right-hand side is then $y\left(\frac{1}{2} + \frac{1}{4} + \frac{1}{8}+\dots\right)$, and the bracket evaluates to 1. Boom.

## But wait a moment

We’ve found *a* solution. How do we know it’s the only one?

What if you notice that $y’’ = y’’’ + y^{(4)} + \dots$ and get to $y = y’ + 2y’’$?

That solves as $y = Ae^{x/2} + Be^{-x}$. Isn’t that a solution?

Actually, not if $B$ is non-zero – the sum of the derivatives doesn’t converge – but all the same, you could (in principle) do this with any of the derivatives and is there any guarantee that there’s no other solution?

## Stand back.

I’m going to try something I haven’t done in 25 years ((I mean, obviously I have, I did it just before writing up the post. But the point stands.)): Laplace transforms.

The tl;dr is that you convert every term of your differential equation (in “$t$-space”, even though we’re really using $x$ here) into a different term in $s$-space, do your magic there and convert back.

Do not worry about what $t$- and $s$- space are, or how the transform works; that’s not the purpose of this article.

The Laplace transform of the ODE:

$0 = -y + y’ + y’’ + y’’’ + \dots$

is

$\begin{aligned}[t]0 = &-Y(s) &

& + sY(s) &- y(0)

& + s^2Y(s) &- sy(0) &- y’(0)

& + s^3Y(s) &- s^2y(0) & -sy’(0) & -y’‘(0)

& + \dots
\end{aligned}$

At first glance, that looks awful. And then you spot that the columns (apart, slightly, from the first) are geometric sequences! This all reduces to

$0=\left(-2 + \frac{s}{1-s}\right)Y(s) - \frac{1}{1-s}\left(y(0) + y’(0) + \dots\right)$

(As long as $s$ is small enough, I guess.)

Now, we don’t really care about the $y(0)$s and so on, and we can just treat them as a single constant. We can also simplify the bracket on the first term:

$0 = \frac{2s-1}{1-s}Y(s) - \frac{1}{1-s}A$

And as long as $s$ isn’t $1$ ((I don’t even know if that makes sense, honestly, like I said, it’s been 25 years)), the solution to that is $Y(s) = \frac{A}{2s-1}$.

The inverse Laplace transform of *that* is $y(x) = Ae^{x/2}$, as we had before.

I’m still a bit uneasy about the convergence issues on $s$, but I’m going to pretend they don’t matter. At any rate, I’ve convinced *myself* that this is the only solution, and that’s the main thing.